Given a list of stringswordsrepresenting an English Dictionary, find the longest word inwordsthat can be built one character at a time by other words inwords. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string.

Example 1:

Input:

words = ["w","wo","wor","worl", "world"]

Output:
 "world"

Explanation:

The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input:

words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]

Output:
 "apple"

Explanation:

Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Note:

All the strings in the input will only contain lowercase letters.

The length of

words

will be in the range

[1, 1000]

.

The length of

words[i]

will be in the range

[1, 30]

.

class Solution {
    public String longestWord(String[] words) {
        Trie trie = new Trie();
        int index = 0;
        for (String word: words) {
            trie.insert(word, ++index); //indexed by 1
        }
        trie.words = words;
        return trie.dfs();
    }
}
class Node {
    char c;
    HashMap<Character, Node> children = new HashMap();
    int end;
    public Node(char c){
        this.c = c;
    }
}

class Trie {
    Node root;
    String[] words;
    public Trie() {
        root = new Node('0');
    }

    public void insert(String word, int index) {
        Node cur = root;
        for (char c: word.toCharArray()) {
            cur.children.putIfAbsent(c, new Node(c));
            cur = cur.children.get(c);
        }
        cur.end = index;
    }

    public String dfs() {
        String ans = "";
        Stack<Node> stack = new Stack();
        stack.push(root);
        while (!stack.empty()) {
            Node node = stack.pop();
            if (node.end > 0 || node == root) {
                if (node != root) {
                    String word = words[node.end - 1];
                    if (word.length() > ans.length() ||
                            word.length() == ans.length() && word.compareTo(ans) < 0) {
                        ans = word;
                    }
                }
                for (Node nei: node.children.values()) {
                    stack.push(nei);
                }
            }
        }
        return ans;
    }
}